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I have a conceptual question about surface gravity for non-spherical bodies. This is partly motivated by curiosity and science fiction, but I'm interested in the actual Newtonian physics.

Consider a hypothetical celestial body with:

  • total mass equal to Earth's,
  • uniform density,
  • shape: a rectangular cuboid with dimensions $L \times L \times H$, where $H \approx 0.2L$ (i.e. a very flat slab),
  • the material is assumed to be rigid enough to maintain this shape against self-gravity.

(Optionally, the body could rotate, but I'm mainly interested in the non-rotating case first.)

Questions

  1. Surface gravity distribution on the large face
    Let $g = 1$ at the center of one of the large ($L \times L$) faces. Let's call this point $C_1$.
    • How does the magnitude and direction of the gravitational field change as one moves toward the edges and corners of that face?
    • Is there a known analytic expression or approximation for this?
  2. Gravity on the smaller faces (the "edges")
    What is the gravitational field like at the center of one of the ($L \times H$) faces? Let's call this point $C_2$.
    • Is the magnitude smaller than that at the center of the large face ($C_1$)?
    • Is the field there approximately perpendicular to the face, or significantly tilted?
  3. Relative magnitudes
    Qualitatively (or quantitatively, if possible), how do the following compare:
    • center of large face
    • edge of large face
    • center of small face
    • corner of the cuboid
  4. Direction of "down"
    Is it correct that, away from the center of the large face ($C_1$), the gravitational field develops a lateral component pointing toward the center of mass?

My understanding so far

For a sphere, the shell theorem allows us to treat the mass as concentrated at the center. For this cuboid, that symmetry is lost, so the field should be obtained by integrating contributions from all mass elements, leading to position-dependent magnitude and direction.

My intuition is that:

  • Near the center of the large face ($C_1$), the field should resemble that of a (finite) plane.
  • Near edges and corners, the field tilts inward and weakens.

What confuses me a lot

At point $C_1$, technically all the mass would be "under my feet", but most of it would actually be around me, and I think that I would experience the net sum of all the sideways vectors. Those would still happen to pull me downwards, perpendicularly to the $L \times L$ face. I think that in that case, the actual gravitational force would be weaker than if the same mass were shaped like a planet.

And if I stand on $C_2$, much more of the mass would straight underneath my feet (I'd experience fewer vectors pulling me sideways, or they'd be less tilted), but then a lot of the mass would also be further away from me, which I believe means that I would experience a weaker pull.

I'd appreciate confirmation, corrections, or pointers to known results (analytical or numerical).

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  • $\begingroup$ An object or Earth's mass and density would be unlikely to maintain a shape like that. There's a reason planets are round and it's due to the fact that the gravitational force on itself will tend to force the objects into a round shape. $\endgroup$ Commented 2 days ago
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    $\begingroup$ What is the conceptual question here? Regarding the relative magnitudes, you just have to do the calculation (possibly numerically), see e.g. this paper: The gravitational field of a cube. The field at C2 has to be perpendicular to the face by symmetry. You can also see a simulation of how the gravitational field looks like for a disc in the following YouTube video by VSauce: Is the Earth actually flat?. $\endgroup$ Commented 2 days ago
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    $\begingroup$ There is no conceptual question here beyond computing the potential and field of a cuboid of constant density. $\endgroup$ Commented 2 days ago
  • $\begingroup$ Also, be careful to remember that the gravitational field will in general not point towards the centre of mass of the object (like in the case of a cube of uniform density; see this blogpost: If the Earth were a cube). Also, refer to this Phy.SE question: Is the force of gravity always directed towards the center of mass? $\endgroup$ Commented 2 days ago

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This does have a closed-form solution, which I will state but not derive. First, define $$f(u,v,w) \equiv uv\operatorname{arsinh}\frac{w}{\sqrt{u^2+v^2}} + vw\operatorname{arsinh}\frac{u}{\sqrt{v^2+w^2}} + wu\operatorname{arsinh}\frac{v}{\sqrt{w^2+u^2}} - \frac{u^2}{2}\arctan\frac{vw}{u\sqrt{u^2+v^2+w^2}} - \frac{v^2}{2}\arctan\frac{wu}{v\sqrt{u^2+v^2+w^2}} - \frac{w^2}{2}\arctan\frac{uv}{w\sqrt{u^2+v^2+w^2}}.$$ The gravitational potential of a cuboid of uniform density $\rho$ centered at the origin, aligned with the Cartesian axes and having dimensions $2a$, $2b$, $2c$ in the $x$, $y$, $z$ directions respectively is $$\phi(x,y,z) = -G\rho \sum_{k_1,k_2,k_3 \in \{-1,+1\}} k_1 k_2 k_3 f(x+k_1 a,y+k_2 b,z+k_3 c) \\ = -G\rho[f(x+a,y+b,z+c) + f(x+a,y-b,z-c) + f(x-a,y+b,z-c) + f(x-a,y-b,z+c) - f(x-a,y+b,z+c) - f(x+a,y-b,z+c) - f(x+a,y+b,z-c) - f(x-a,y-b,z-c)]$$ and the gravitational field is $-\nabla\phi$. Sample plots are shown below:

  • 3D gravitational field 1

  • Larger, square face (i.e. top and bottom)

    • Normal component: 2
    • Tangential component: 3

  • Smaller, square face (i.e. sides)

    • Normal component: 4
    • Tangential component: 5

So your intuition that the gravitational field away from the center of a face has a tangential component is correct. However, for your dimensions, the gravitational field at the center of a larger, square face is actually about 8% weaker than that at the center of a smaller, rectangular face. In addition, within each of the larger, square faces, the point with the strongest magnitude of gravity is not the center, but rather near the center of each of the four edges.

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    $\begingroup$ Note that if one asks only about points which are several thicknesses inside from any edge, the Gauss's Law assumptions more commonly associated with an infinite charged sheet start to apply and you get a(n approximately) uniform field perpendicular to the surface. But $H≈0.2L$ is not especially thin. $\endgroup$ Commented yesterday
  • $\begingroup$ My congratulations. Excellent answer. $\endgroup$ Commented 15 hours ago
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This isn't really an answer, but if you want a quantitative analysis, just do it on a computer. It's a Cartesian coordinate analysis with three (total) nested loops.

The only difficulty is keeping track of a vector summation for the output.

For a more qualitative inspection, searching "gravity of a cube" produces myriad results (can you imagine if Newton and everybody had an Internet connection?):

Enter image description here

So give it a shot, and get back to us.

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